Quadratic EquationHard
Question
The set of all real numbers x for which x2 - |x + 2| + x > 0 is
Options
A.(- ∞, - 2) 
(2, ∞)
(2, ∞)B.(- ∞, - √2) (√2, ∞)
(√2, ∞)C.(- ∞, - 1) (1, ∞)
(1, ∞)D.(√2, ∞)
Solution
Given, x2 - | x + 2 | + x > 0 ......(i)
Case I When x + 2 ≥ 0
∴ x2 - x - 2 + x > 0
⇒ x2 - 2 > 0
⇒ x < - √2 or x > √2
⇒ x ∈ (- 2, -√2) (√2, ∞) ......(ii)
Case II When x + 2 < 0
∴ x2 + x + 2 + x > 0
⇒ x2 + 2x + 2 > 0
⇒ (x +1)2 + 1 > 0
which is true for all x.
∴ x ≤ - 2 or x ∈ (- ∞, - 2) ......(ii)
From Eqs. (ii) and (iii), we get
x ∈ (- ∞, - √2) (√2, ∞)
Case I When x + 2 ≥ 0
∴ x2 - x - 2 + x > 0
⇒ x2 - 2 > 0
⇒ x < - √2 or x > √2
⇒ x ∈ (- 2, -√2)
(√2, ∞) ......(ii)Case II When x + 2 < 0
∴ x2 + x + 2 + x > 0
⇒ x2 + 2x + 2 > 0
⇒ (x +1)2 + 1 > 0
which is true for all x.
∴ x ≤ - 2 or x ∈ (- ∞, - 2) ......(ii)
From Eqs. (ii) and (iii), we get
x ∈ (- ∞, - √2)
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